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Contents

1. Foreword i

2. How to use this booklet 1

3. Study and examination tips 23.1 Tips for Physical Sciences Paper 1 2

3.2 Tips for Momentum and Impulse 4

4. Overview of Momentum and Impulse 5

5. MOMENTUM AND IMPULSE 6 5.1 Momentum 6 5.2 IMPULSE AND CHANGE IN MOMENTUM 7 5.3 Newton’s Second Law in terms of momentum 8 5.4 Impulse 9 5.5 The Principle of Conservation of Linear Momentum 10

6. Check your answers 18

7. Message to Grade 12 learners from the writers 26

8. Thank you and acknowledgements 26

1

2. HOW TO USE THIS BOOKLET • This book is intended to guide you through the topics. It is a complimentary publication

to the textbooks, worksheets and other learning materials you might have. Your efforts

to practice and master the concepts outlined will assist you to develop confidence in

the learning and assessments you will do.

• Ensure you understand all the relevant concepts, formulae, etc.

• Do extra research on your own to obtain more information on the topics in this booklet.

It is important that you share and discuss your understanding of your findings with your

fellow learners.

• Work through the examples given under each topic, master them and try to provide

more examples yourself. You must go through examples in your textbooks. At the

end of Unit 5, there are exercises for you to do. Work through them on your own and

compare your answers with the answers provided in Unit 6.

• When doing calculations, take note of all the information given. It is intended to help

you. Read the given information with the intention of understanding what you must do

with it. Always look for the action verbs in the instruction, in order to answer correctly.

Examples include: determine, explain, discuss, calculate, list, compare.

• If you are given measurements in millimetres (mm), convert these into meters.

• Familiarise yourself with the scientific notations used, e.g. kilo, mega, giga and write

them as 10x.

• It is important to show SI units on the final answer. This will earn you a mark.

• At the beginning of each topic section, an extract from the Examination Guidelines is

included. You must use this to check what you are expected to master. Revise what

you have not mastered and do more exercises.

2

3. STUDY AND EXAMINATION TIPS Some tips on how to approach the examinations:

3.1 Tips for Physical Sciences Paper 1

Format of question papers PAPER QUESTION TYPE DURATION TOTAL

1• 10 Multiple-choice questions – 20 marks

• Structured questions – 130 marks3 hours 150

2• 10 Multiple-choice questions – 20 marks

• Structured questions – 130 marks3 hours 150

Weighting of prescribed content

P1 CONTENT P2 CONTENT

Content Marks % Content Marks %Mechanics 108 72 Organic molecules 54 36

Electricity and Magnetism 42 28 Electrochemistry 42 28

Light 54 36

Laws, definitions and principles

Study and understand the laws, definitions, concepts and principles stated in the examination

guidelines. Careful attention must be paid to the key words.

Using formulae • Always use formulae from the formula sheet.

• Common mistakes occur when you change the subject of the formula; thus, it is

recommended that you substitute into the given formula and isolate the unknown.

However, those of you who are confident with changing the subject of the formula may

continue to use this method as it is an efficient method.

• Familiarise yourself with the formula sheet to avoid wasting time during the examination.

• Round off your final answer to a minimum of two decimal places and include the correct

unit.

3

Answering multiple-choice questionsMultiple-choice questions (MCQs) ask a learner to recognise a correct answer among a set

of options that include 3 or 4 wrong answers (called distracters).

The level of difficulty will vary in MCQs. Resorting to guessing may lead to giving incorrect

answers.

There are strategies that can be used to maximise your success in answering MCQs. The

best way to improve your chances, of course, is to study carefully before the exam and to

make sure that you understand your work, instead of just memorising information.

Here are a few tips to help reduce the possibility of making mistakes or of getting confused

by distracters that look similar to the correct answer.

Step 1 • Always cover the possible options given with a piece of paper or your hand while you

read the stem or body of the question.

• Read carefully and make sure you understand what you are required to do. Rushing

through a question may result in misinterpretation.

Step 2 Try to anticipate the correct answer before looking at the given options.

Step 3 • Uncover the responses and read them. If you see the response you anticipated or

one that closely matches your anticipated response, circle/ mark it and then check the

others to make sure none of them is a better response. It is important to read all given

responses.

• If your anticipated response is not amongst the given ones, or if you are not able to

anticipate an answer, read the given options and eliminate those you know are wrong.

• By eliminating wrong options, you will be left with fewer options from which to select

your answer. This makes it easier to look for the correct option.

Step 4 • Look at the remaining options; compare them for differences and then refer to the stem

of the question to find the correct answer.

If you cannot answer a question in 1 minute, skip it and plan to come back to it later.

4

3.2 Tips for Momentum and Impulse

• The topic of Momentum & Impulse is assessed in Paper 1.

• Choose a sign convention and indicate it on your answer sheet, in your answer to the

question on momentum and impulse.

• Copy the formula as it is written in the data sheet, without attempting to manipulate it.

• Ensure that all values are in SI units, i.e. carry out the necessary conversion.

• Show all substitutions.

• Include the correct unit for the answer.

Sketching of force and free-body diagrams

• You must be able to distinguish between a free-body and force diagrams.

• The mark allocation is a guide to the number of forces required in the diagram, e.g. for

4 marks, 4 labelled forces should be drawn.

• Extra forces, leaving out the arrowheads, and arrows that do not touch the dot (object)

will result in lost marks, due to negative marking.

• Label the forces correctly.

• DEFINITION OF TERMS – know and understand all the definitions and terms

LAW/CONCEPT/TERM DEFINITION

Momentum The product of an object's mass and its

velocity

Newton's Second Law of Motion in terms of

momentum

The net (or resultant) force acting on an

object is equal to the rate of change of

momentum of the object in the direction of

the net force.

Impulse

The product of the net force acting on an

object and the time the net force acts on

the object.

An isolated system (in Physics):One on which the net external force acting

on the system is zero.

The principle of conservation of linear

momentum

The total linear momentum of an isolated

system remains constant (is conserved) in

magnitude and direction.

5

4.

OVE

RVI

EW O

F M

OM

ENTU

M A

ND

IMPU

LSE

Mo

men

tum

(p

) of

an

obje

ct is

the

prod

uct o

f its

mas

s (m

) an

d ve

loci

ty (

v).

The

mas

s of

the

obje

ct m

ust b

e in

kilo

gram

s

(kg)

.

The

vel

ocity

of t

he o

bjec

t mus

t be

in (

m·s

-1)

For

mul

a: p

= m

vS

I uni

t : kg

·m·s

-1

Mom

entu

m (

p) is

a v

ecto

r, th

us d

irect

ion

is

impo

rtan

t.

Mom

entu

m o

f an

obje

ct d

epen

ds o

n its

:

1.

mas

s

2.

velo

city

Ch

ang

e in

mo

men

tum

(∆p

) is

•the

differenceb

etweenfi

nalm

omentum

(pf)

and

(pi)

• a

vec

tor

quan

tity

(dire

ctio

n is

impo

rtan

t)

• de

pend

ent o

n ∆v

SI u

nit:

kg·m

·s-1

MO

ME

NT

UM

AN

D IM

PU

LS

E

For

mul

a:

Δp =

p f – p i

Δp =

m·Δ

vΔp

= m

(vf –

v i)

Law

of c

onse

rvat

ion

of li

near

mom

entu

m:

The

tota

l lin

ear

mom

entu

m o

f an

isol

ated

syst

em r

emai

ns c

onst

ant (

is c

onse

rved

) in

mag

nitu

de a

nd d

irect

ion.

Σpb

efo

re =

Σpaf

ter

New

ton

’s s

eco

nd

law

in t

erm

s o

f lin

ear

mo

men

tum

: th

e ne

t (re

sulta

nt)

forc

e ac

ting

on a

n ob

ject

is e

qual

to th

e ra

te o

f cha

nge

of

mom

entu

m o

f the

obj

ect i

n th

e di

rect

ion

of th

e

net f

orce

.

Fne

t = ∆t

∆p

, the

refo

re

F net∆t

= ∆

pIm

pu

lse

is th

e pr

oduc

t of a

net

forc

e an

d tim

e

(Fne

t∆t)andis

Fo

rmu

lae:

Fne

tΔt =

Δp

F netΔt

= m

(vf –

v i)Im

puls

e is

a v

ecto

r qu

antit

y, d

irect

ion

is

impo

rtan

t.

SI u

nit

s: 1

N·s =

1 kg

·m·s

-1

Type

s of

col

lisio

ns

•

Ela

stic

- to

tal k

inet

ic e

nerg

y is

con

serv

ed:

ΣE

ki = Σ

Ekf

•

Inel

astic

– to

tal k

inet

ic e

nerg

y is

not

con

serv

ed.

ΣE

ki ≠

ΣEkf

To

tal l

inea

r m

omen

tum

is c

onse

rved

in

bo

th c

ollis

ions

.

6

5. MOMENTUM AND IMPULSE5.1 Momentum

MOMENTUM • Define momentum as the product of an object’s mass and its velocity. • Describe linear momentum as a vector quantity with the same direction as the velocity

of the object. • Calculate the momentum of a moving object using p = mv, in the context of technology.

NEWTON’S SECOND LAW IN TERMS OF MOMENTUM • State Newton’s Second Law of Motion in terms of momentum: • The net (or resultant) force acting on an object is equal to the rate of change of

momentum of the object in the direction of the net force.

• Net force is equal to the rate of change in momentum, Fnet

= ∆t ∆p

.

• Calculate the change in momentum when a resultant force acts on an object and its velocity:

– increases in the direction of motion, e.g. 2nd stage rocket engine fires – decreases, e.g. brakes are applied – reverses its direction of motion, e.g. a soccer ball is kicked back in the direction it

came from.

IMPULSE • Define impulse as the product of the net force acting on an object and the time that the

net force acts on the object. • Impulse is equal to the change in momentum. • Use the impulse-momentum theorem provided below to calculate: the force exerted; the

length of time the force is applied; the change in momentum for a variety of situations involving the motion of an object in one dimension, in the context of technology.

Fnet

Δt = m Δv, • Explain how the concept of impulse applies to road safety considerations in everyday

life, e.g. airbags, seatbelts, crumple zones and arrestor beds.

CONSERVATION OF MOMENTUM AND ELASTIC AND INELASTIC COLLISIONS

• Explain what is meant by: – An isolated system (in Physics): An isolated system is one in which the net

external force acting on the system is zero. – Internal and external forces. • State the principle of conservation of linear momentum: The total linear momentum of

an isolated system remains constant (is conserved) in magnitude and direction. • Use the conservation of momentum principle to solve problems involving momentum

in the context of technology. Restrict problems to two objects moving in one dimension (along a straight line) with the aid of an appropriate sign convention.

• Distinguish between elastic collisions and inelastic collisions by calculation.

7

5.2 IMPULSE AND CHANGE IN MOMENTUM Momentum and impulse

• Momentum is the product of an object’s mass and its velocity.

• Momentum = mass x velocity

p = m v

• Since mass is measured in kg and velocity is measured in m.s-1, the S.I unit for

momentum is kg.m.s-1

• Momentum is a vector quantity, i.e. it has both magnitude and direction.

• There are two types of momentum, viz. angular momentum (i.e. due to the rotation of

an object), and linear momentum (movement in a straight line). We are only concerned

with linear momentum. • Impulse is the product of the net force acting on an object and the time the net force

acts on that object.

• In mathematical terms, it is expressed in symbols as: Impulse = Fnet

Δt

• Impulse – Momentum Theorem: Fnet

Δt = Δp

WORKED EXAMPLES

a) Determine the momentum of a 6 kg object moving at 5 m.s-1 to the right.

Solution p = mv

= 6(5)

= 30 kg.m.s-1 to the right

b) A car has a momentum of 20 000 kg·m·s–1. Calculate the car’s new momentum if

its velocity is doubled.

Solution p = mv

= m(2v)

= 2(mv)

= (2)(20 000)

= 40000 kg.m.s-1

The momentum will double and will be equal to 40 000 kg·m·s-1 in the same

direction as before.

8

(c) Calculate the velocity of the car if its momentum is 60 000 kg·m·s–1 and the

mass is 2 000 kg?

Solution p = mv

60 000 = 2 000 v

v = 30 m·s-1 in the same direction as the momentum.

5.3 Newton’s Second Law in terms of momentum

• The net (or resultant) force acting on an object is equal to the rate of change of

momentum of the object in the direction of the net force, Fnet

Δt = Δp

• The momentum of an object changes with its velocity.

• Change in momentum = mass (final velocity – initial velocity)

p = pfinal

- pinitial

= mvf - mv

i

= m (vf - v

i)

ExampleA 500 g soccer ball is kicked at 12 m·s-1 towards a wall. It rebounds off the wall at 2 m·s-1.

Calculate the change in momentum.

Solution

Choose initial direction as the positive.

pi = mv

i = 0,5 × 12 = 6 kg·m·s-1

pf = mv

f = 0,5 × –2 = – 1 kg·m·s-1

Δp = pf – p

i = –1 – 6 = – 7 kg·m·s-1

Δp= 7 kg·m·s-1 away from the wall

9

5.4 IMPULSE

Example 1A golf ball with a mass of 0, 1 kg is driven from the tee.

The golf ball experiences a force of 1000 N while in contact with the golf club and moves

away from the golf club at 30 m·s-1. Calculate the time during which the golf club was in

contact with the ball.

Fnet

Δt = Δp

1000(Δt) = (30 − 0)

t = 3 × 10−3 s

Example 2The following graph shows the force exerted on a hockey ball over time. The hockey ball is

initially stationary and has a mass of 150 g.

Calculate the magnitude of the impulse of the hockey ball.

Solution F

netΔt = impulse = area under graph

= 21

base x height

= 21

(0,5 x 150)

= 37,5 N.s

10

TEST YOUR KNOWLEDGE Question 1

A 12 kg object speeds up from an initial velocity of 10 m∙s-1 north to a final velocity of 15 m.s-1

north. Calculate the change in momentum.

Question 2

A 20 kg object slows down from an initial velocity of 40 m∙s-1 west to a final velocity of 5 m∙s-1

west. Calculate the change in momentum.

Question 3

A car with a mass of 1 200 kg is travelling at 72 km.h-1 when the driver sees a dog running

across the street. He applies the brakes and 10 seconds later the car is moving at 18 km.h-1.

I. Calculate the initial momentum of the car.

II. Calculate the final momentum of the car.

III. Calculate the change in momentum of the car.

IV. Calculate the net force applied by the brakes of the car.

5.5 THE PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM

• The total linear momentum of an isolated system remains constant (is conserved) in


• An isolated system is one in which the net external force acting on the system is zero.

COLLISIONS • There are a number of possible types of collision that can take place between two

objects moving along the same line (one dimension), e.g.:

– one moving object collides with a stationary object

– two objects move towards each other and collide

– two objects move in the same direction and collide

– one unit separates into parts (explosion) e.g. an exploding bomb:

Possible results from collisions: • both objects collide and stop

• one object moves and the other object stops moving

• both objects move but are not joined

• two objects join together and move off at the same speed

11

Elastic and Inelastic Collisions:

• In an elastic collision, total kinetic energy and total linear momentum are

conserved.

• In an inelastic collision, total linear momentum is conserved, but not total kinetic

energy.

• The conservation of kinetic energy is determined by calculating the total kinetic

energy of all parts of the closed system, before the collision, and then comparing

that to the total kinetic energy of all the parts of the closed system after the

collision.

WORKED EXAMPLESA tennis ball with a mass of 0,06 kg is thrown horizontally towards a wall. It hits the wall at

5 m∙s -1, and bounces back at 3 m∙s-1.

Calculate:

i. The initial momentum of the tennis ball.

ii. The final momentum of the tennis ball.

iii. The change in momentum of the tennis ball.

SolutionTake the direction to the right as being positive.

i. Initial momentum: pi = mv

i

= (0,06) (5)

= +0,3 kg∙m∙ s-1

= 0,3 kg∙m∙ s-1 to the right

ii. Final momentum: pf = mv

f

= (0.06)(-3)

= -0,18 kg∙m∙s-1

= 0,18 kg∙m∙s-1 to the left

iii. Change in momentum = pf - p

i

= (-0,18 – 0.3)

= -0,48 kg∙m∙s-1

= 0,48 kg∙m∙s-1 to the left

12

TEST YOUR KNOWLEDGE Question 1

A 35 kg girl is standing near and to the left of a 43 kg boy on the frictionless surface of a

frozen pond. The boy throws a 0, 75 kg ice ball to the girl with a horizontal speed of 6.2

m·s-1. What are the velocities of the boy and the girl immediately after the girl catches the

ice ball?

Question 2

Two billiard balls (ball 1 and ball 2),each with a mass of 150 g, collide head-on, as shown in

the diagram below. Ball 1 was travelling at a speed of 2 m·s-1 and ball 2 was travelling at a

speed of 1, 5 m·s-1. After the collision, ball 1 travels away from ball 2 at a speed of 1, 5 m·s-1.

I. Calculate the velocity of ball 2 after the collision.

II. Prove that the collision was elastic. Show by means of calculations.

EXERCISES QUESTION 1

1.1 To improve the safety of motorists, modern cars are built so that the front-end crumples

upon impact. A 1200 kg car is travelling at a constant velocity of 8 m∙s-1. It hits a wall

and comes to a complete stop in 0, 25 s.

I. Calculate the change in momentum.

II. Calculate the average net force exerted on the car.

III. For the same impulse, determine the average net force exerted on the car, if it had a

rigid bumper and frame that stopped the car in 0,040 s.

13

1.2 A rubber ball with a mass of 800 g is dropped and hits the floor with an initial downward

velocity of 6 m∙s-1. The ball bounces back with a final velocity of 4 m∙ s-1.

Calculate the change in momentum of the ball.

QUESTION 2

Car A that has a mass of 2000 kg is moving at 8 m∙s-1 to the west. It collides with Car B that

has a mass of 4000 kg moving at 1 m∙s-1 to the east.

2.1 State the principle of conservation of linear momentum.

2.2 Calculate the change in momentum of the 2000 kg car.

2.3 Calculate the velocity of car B, if car A moves with a velocity of 4 m.s-1

After the collision.

14

QUESTION 3

A cannon has a mass of 1 250 kg. It fires a cannon-ball during a routine exercise. The

cannon is 1000 times heavier than the cannon ball. The cannon ball leaves the barrel at a

horizontal velocity of 70 m.s–1.

The cannon comes to rest 1 second after the cannon-ball was fired.

3.1 Define, in words, the term impulse.

3.2 Calculate the following:

3.2.1 maximum velocity with which the cannon moves backwards;

3.2.2 magnitude of the average net force that causes the cannon to come to rest.

QUESTION 4

Collisions happen daily on the roads in our country. In one of these collisions, a car with a

mass of 1 600 kg is travelling at a speed of 30 m·s-1 to the left. It collides head-on with a

minibus that has a mass of 3 000 kg, which is travelling at 20 m·s-1 to the right. The two

vehicles move together as a unit in a straight line after the collision.

4.1 Calculate the velocity of the two vehicles after the collision.

4.2 Do the necessary calculations to show that the collision was inelastic.

4.3 The billboard below advertises a car from a certain manufacturer.

Use your knowledge of momentum and impulse to justify how the safety features mentioned

in the advertisement contribute to the safety of the passengers.

15

QUESTION 5

The diagram below shows two trolleys (P and Q) that are held together by means of a

compressed spring on a flat, frictionless horizontal track. The mass of P and Q is 300 g and

500 g, respectively.

When the trolleys are released, it takes 0,5 s for the spring to unwind to its natural length.

Trolley Q then moves to the right at 2 m∙s-1.

5.1 State the principle of conservation of linear momentum in words.

5.2 Calculate the following:

5.2.1 velocity of trolley P after the trolleys are released;

5.2.2 magnitude of the average force exerted by the spring on trolley Q.

5.3 Is this an elastic or inelastic collision?

16

QUESTION 6

The diagram below shows a bullet with a mass of 20 g that is travelling horizontally. The

bullet strikes a stationary 7 kg block and becomes embedded in it. The bullet and the block

travel on a rough horizontal surface for a distance of 2 m before coming to a stop.

At the moment of impact, the bullet and the block begin to move with a velocity of 2,39 m∙s-1.

6.1 State the Principle of Conservation of Linear Momentum in words.

6.2 Calculate the magnitude of the velocity with which the bullet hits the block.

QUESTION 7

A trolley with a mass of 5 kg moves at 4 m·s-1 east across a frictionless horizontal surface.

A brick with a mass of 1,5 kg is dropped onto the trolley.

7.1 State, in words, the Law of Conservation of Linear Momentum.

7.2 State the condition for an elastic collision.

7.3 Calculate the change in momentum of the 5 kg trolley.

17

QUESTION 8

A girl with a mass of 65 kg is driving a 535 kg snowmobile at a constant velocity of 11,5 m∙s-1

[60o E of N]. Calculate the momentum of the girl-snowmobile system.

QUESTION 9

A boy on a skateboard moves to the right at constant velocity. The joint mass of the boy

and skateboard is 50 kg. He catches a ball with of mass 0,4 kg while the ball is travelling

horizontally to the left at a speed of 6 m·s-1. After the boy catches the ball, they both move

to the right at 1,49 m·s-1.

9. 1 Define the term impulse.

9.2 Calculate the magnitude of the average force that the boy exerts on the ball when he

catches it, if he and the ball exert a force on each other for a period of 0,1 s.

9.3 Write down the Principle of Conservation of Linear Momentum in words.

9.4 Calculate the magnitude of the velocity (v) of the boy before he catches the ball.

9.5 Prove that this is an inelastic collision and show all the necessary calculations.

18

6. CHECK YOUR ANSWERS TEST YOUR KNOWLEDGE SOLUTIONS

MOMENTUM AND IMPULSE

Question 1

Take north as positive.

∆p = m(vf - v

i )

= 12(15-10)

= 60 kg.m.s-1, north

Question 2

Take west as positive.

∆p = m(vf - v

i )

= 20(5-40)

= -700 kg.m.s-1

=700 kg.m.s-1 west

Question 3

I. p = mv

= 1200(20)

= 24000 kg.m.s-1 forward

II. p = mv

=1200(5)

=6000 kg.m.s-1 forward

III. ∆p = m(vf - v

i )

=1200(6000-24000)

= 2,16 x 107 kg.m.s-1 in the opposite direction

IV. Fnet

∆t = m∆v

Fnet

(10) = -2,16x 107

Fnet

= 2,16 X 106 N in the opposite direction

19

CONSERVATION OF MOMENTUM

Question 1

mBv

B + m

P v

P = m

Bv

B + m

Pv

P

0 + 0 = 43 (vB) + 0,75(6.2)

vB = 0,108 m.s-1

mGv

G + m

Pv

P = v(m

G +m

P)

35(0) + 0,75(-6,2) = v(35,75)

v = - 0,13 m.s-1

v = 0, 13 m.s-1 to the left

Question 2

I. Σpbefore

= Σpafter

m1v

1i + m

2v

2i = m

2v

1f + m

2v

2f

(0,15)(2)+(0,15)(-1,5) = (0,15)(-1,5)+0,15vf2

0,3-0,225 = -0,225+0,15v2

vf2 = 3 m∙s-1 to the right.

II. ΣEK(before)

= 2

m1V

i21

+ 2

m2v

i22 =

2

(0,15)(2)2

+ 2

(0,15)(–1,5)2

=0,469 J

ΣEK(after)

=2

m1V

f21

+ 2

m2v

f22 =

2

(0,15)(–1,5)2

+ 2

(0,15)(2)2

=0,469 J

So, ΣEK(before)

= ΣEK(after)

∴ Collision is elastic.

20

SOLUTIONS TO EXERCISES

QUESTION 1

1.1

I. ∆p = m(vf - v

i )

= (1200 )(0 - (8)

= (1200)(8.0)

= 9.6 ×103 kg∙m∙s-1 (west)

II. Fnet

∆t = -9,6 ×103 N∙s

Fnet

= ∆t –9,6 X 103

Fnet

= 0,25

–9,6 X 103

= -3,8 x 104 N

= 3 X 104 N West

III. Fnet

= ∆t –9,6 X 103

Fnet

= 0,040

–9,6 X 103

= -2,4 x 105

= 2,4 x 105 N West

1.2 ∆p = m(vf - v

i)

= 0,8 (- 4 – 6)

= 8 kg·m·s-1 upwards

QUESTION 2

2.1 The total linear momentum of an isolated system remains constant in magnitude and

direction.

2.2 ∆p = m (vf - v

i) (taking east as positive)

= 2000[4 –(-8)]

=24000 kg.m.s-1 east

2.3 Σpi = Σp

f

m1v

i1 + m

2 v

i 2 = m

1v

f1 + m

2 v

f2

(4000)(1) + 2000(-8) = 4000vf+ (2000) (4)

vf = -5 m.s-1

vf = 5 m.s-1 west

21

QUESTION 3

3.1 Impulse is the product of the resultant/ net force acting on an object and the length of time

that the resultant/ net force acts on the object.

3.2

3.2.1

OPTION 1 OPTION 2

Take the direction towards the left as positive. Take the direction towards the right as positive.

Σpi = Σp

f Σp

i = Σp

f

0 = mvcannon

+ mvball

0 = mvcannon

+ mvball

0 = (1250)v + 1,25 (–80) 0 = (1250)v + 1,25 (80)

v = 100 / 1250 = 0,084 m.s–1, left v = 100 ÷ 1250 = – 0,08 m.s–1

v = 0,084 m.s–1, left

3.2.2

Take the direction towards the left as positive.

Fnet

∆t = m∆v = mvf – mv

i

Fnet

(1) = 1250 (0 – (-0,084))

Fnet

= 105 N

OR

vf = v

i + a∆t

0 = -0,084 + a(1)

a = 0,084 m.s-2

Fnet

= ma

= (1250)(0,084)

= 105 N to the right

22

QUESTION 4

4.1 Consider the motion to the right as positive.

Σpi = Σp

f

m1v

i1 + m

2v

i2 = (m

1 + m

2)v

f

(1 600)(-30) + (3 000)(20) = (1 600 + 3 000) vf

-48 000 + 60 000 = (4 600)vf

vf = 2,6 m·s-1 to the right

4.2

Before collision:ΣE

k=

2

1 m

1V

i21 +

2

1 m

2V

i22 =

2

1 (1 600)(-30)2 +

2

1 (3 000)(20)2

= 720 000 + 600 000

= 1,32 x 106 J

After collision:ΣE

k =

2

1 m

1V

f21 +

2

1 m

2V

f22 =

2

1 (1 600 + 3 000)(2,6)2

= 15548 J

ΣEk before collision is not equal to ΣE

k after collision; therefore, the collision is inelastic.

4.3 During a collision, the crumple zone / airbag increases the time during which momentum

changes. So, according to the equation …

Fnet

= ∆t ∆p

, the force during impact will decrease.

QUESTION 5

5.1 The total linear momentum of an isolated system remains constant in magnitude and

direction.

5.2.1

∑pi = ∑pf

m1v1

i + m

2v

2i = m

1v

1f + m

2v

2f

(m1 +m

2)v

i = m

1v

1f + m

2v

2f

0 = (0,3)v1f + 0,5 (2)

v1f

= -3,33 m∙s-1

= 3,33 m∙s-1 to the left

change in velocity

23

5.2.2

OPTION 1 m(v

f – v

i) = F

net∆t

0,5(2 – 0) = Fnet

(0,5)

Fnet

= 2 N to the right

OPTION 2 m(v

f – v

i) = F

net∆t

0,3(-3,33 – 0) = Fnet

(0,5)

Fnet

=-2 N

Fnet

= 2 N to the right

5.3 Inelastic

QUESTION 6

6.1 The total linear momentum of an (isolated) closed system remains

constant in magnitude and direction.

6.2 Σpbefore

= Σpafter

m1v

i1 + m

2v

I2 = (m

1 + m

2 )v

f

(0,02 )vi1 + (7)(0) = (7,02)(2,39)

vi1 = 838,89 m.s–1

24

QUESTION 7

7.1 The total linear momentum of an isolated system remains constant (is conserved) in


7.2 The total kinetic energy remains constant.

OR

The total kinetic energy before the collision equals the total kinetic energy after the collision.

7.3 Choose the direction to the right as positive.

Σpbefore

= Σpafter

m1v

i1 + m

2 v

i2 = (m1 + m

2)v

f

(5)( 4 ) + (1,5)(0) = (6,5)vf

vf = 3,077 m•s−1

∆p = m(vf - v

i)

= 5(3,077 – 4)

= ‒ 4,62 kg·m·s-1

= 4,62 kg·m·s-1, west

QUESTION 8

6.90 x103 kgm∙s-1 [60o E of N]

QUESTION 9

9.1 The product of the net force and the time that the net force acts on an object.

9.2 Take the motion to the right as positive.

Fnet

∆t = m∆v = m(vf - v

i)

Fnet

(0,1) = 0,4(1,49 - (-6))

Fnet

= 29,96 N

25

9.3 The total linear momentum of a closed system remains constant (is conserved).

9.4 Take the motion to the right as positive.

Σpbefore

= Σpafter

m1v

1i + m

2v

i2 = (m

1 + m

2)v

50vi1 + (0,4)(-6) = 50,4(1,49)

vi1 = 1,55 m∙s-1, to the right

9.5 Total kinetic energy before collision:

ΣEk before

= 21

m1v

i12 +

21

m2v

i22

= (0,5)(50)(1,55)2 + (0,5)(0,4)(-6)2

= 67,26 J

Total kinetic energy after collision:

ΣEk after

= 21

m1v

f12 +

21

m2v

f22

= 21

(m1 + m

2)v2

= (0,5)(50,4)(1,49)2

= 55,95 J

ΣEk before

≠ ΣEk after

∴ inelastic collision

26

7. MESSAGE TO GRADE 12 LEARNERS FROM THE WRITERSTo be successful in life, you need to go the extra mile, make sure you study, and work hard. I

would like to encourage you by referring to Dr John M Tibane who, in his book entitled Turbo

Think, makes the following statements:

“Always do more than is required of you”.

“The number one reason why most people are not wealthy is that they have not decided to

be wealthy”.

To be successful, you need to decide to be successful and do more than is required of you

in order to expedite your success.

8. THANK YOU AND ACKNOWLEDGEMENTS The Technical Sciences Revision Booklet on Momentum was developed by Ms Lerato

Mphachoe, Ms Balungile Manqele (lead writer), Mr Nathi Ndlovu (lead writer), Mr Chele N

Lenkoe and Mr Themba Ncube (Subject Specialists, PED). Special acknowledgement must

be made of Ms JSK Maharaj (Veena), the DBE curriculum specialist who co-ordinated and

finalised the process, in addition to her contributing to the development of the booklet.

These officials contributed their knowledge and experience, and in some cases unpublished

work, which they have gathered over the years, to the development of this resource. The

Department of Basic Education (DBE) gratefully acknowledges these officials for giving up

their valuable time to develop this resource for the children of our country.

Administrative and logistical support was provided by Mr Esrom Fourie. This official was

instrumental in the smooth and efficient management of the logistical processes involved in

this project.

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